Question #114120
Given the vertex and focus of a parabola, (4,2) and (4,5). Determine the following
a. General equation
b. Standard equation
c. Location of the opening
d. Location of the directrix (coordinate)
1
Expert's answer
2020-05-06T19:24:03-0400

Given the vertex and focus of a parabola, (4,2) and (4,5).

Since the x-coordinates of the vertex and focus are the same, so this is a regular vertical parabola, where the x part is squared. Since the vertex is below the focus, this is a right-side up parabola and p is positive. Since the vertex and focus are 5 – 2 = 3 units apart, then p = 3.

So the equation of parabola, we get

(xh)2=4p(yk)(x-h)^2 = 4p(y-k).

So, (x4)2=12(y2)(x-4)^2 = 12(y-2).

Directrix equation is y = y-coordinate of vertex - p

So Directrix equation is y=23=1y = 2 - 3 = -1.


a) We found given parabola is

(x4)2=12(y2)(x-4)^2=12(y-2)

So, General equation of parabola is: x28x12y+40=0x^2 - 8x - 12y + 40 = 0 .


b) Standard equation of parabola is

(x4)2=12(y2)(x-4)^2=12(y-2).


c) Since the vertex is below the focus, this is a right-side up parabola.


d) Location of the Directrix: y=1.y = -1.


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Comments

Assignment Expert
08.05.20, 17:55

Dear Randal Rodriguez, please use the panel for submitting new questions.

Randal Rodriguez
08.05.20, 06:20

3x - 4y + 3 = 0 6x - 8y +7 = 0 a. Find the distance between the two lines. b. Find the equation of the perpendicular line passing through. (-6,4) c. Determine the distance of 2 given equations to point (-6.4)

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